How do you solve ${log}_{5} (4 x + 3) < {log}_{5} 11$ $log_5 (4x+3)<log_5 11$ ?

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Answer 1 · 2017-12-15T16:42:14

Solution : $- 0.75 < x < 2 or x ∣ (- 0.75, 2)$ $-0.75 < x < 2 or x| (-0.75 ,2)$

${log}_{5} (4 x + 3) < {log}_{5} 11$ $log_5 (4x+3) < log_5 11$

$4x+3 >0 :. 4x > -3 or x > -3/4 or x > -0.75$

$4x+3 < 11 or 4x <11-3 or 4x <8$

$or x <2$ . Solution : $-0.75 < x < 2$

In interval notation : $x| (-0.75 ,2)$

Answer 2 · 2017-12-15T19:31:56

Graphical interpritation:

I wanted to provide a graphical approach and interpritation:

Desmos

How do you solve log_5 (4x+3)<log_5 11log5(4x+3)<log511log_5 (4x+3)<log_5 11?